Login/Sign Up
Home » An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500L respectively. The distances (in km) between the depots and the petrol pumps is given in the following table. Assuming that the transportation cost of 10 litres of oil is Rs 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?
An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500L respectively. The distances (in km) between the depots and the petrol pumps is given in the following table. Assuming that the transportation cost of 10 litres of oil is Rs 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?
CBSE | Class 12 | Linear Programming | Maths | Miscellaneous | NCERT
An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500L respectively. The distances (in km) between the depots and the petrol pumps is given in the following table. Assuming that the transportation cost of 10 litres of oil is Rs 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?
Distance in (km)
From / ToAB
D73
E64
F32

Let x and y liters of oil be provided from A to the petroleum siphons, D and E. Along these lines, (7000 – x – y) will be provided from A to petroleum siphon F.

The prerequisite at petroleum siphon D is 4500L. Since x L are moved from warehouse A, the excess (4500 – x) L will be moved from petroleum siphon B.

Also, (3000 – y)L and 3500 – (7000 – x – y) = (x + y – 3500) L will be shipped from stop B to petroleum siphon E and F separately.

The given issue can be addressed diagrammatically as given beneath

NCERT Solutions Mathematics Class 12 Chapter 12 - 41

x ≥ 0, y ≥ 0, and (7000 – x – y) ≥ 0

Then, at that point, x ≥ 0, y ≥ 0, and x + y ≤ 7000

4500 – x ≥ 0, 3000 – y ≥ 0, and x + y – 3500 ≥ 0

Then, at that point, x ≤ 4500, y ≤ 3000, and x + y ≥ 3500

Cost of shipping 10L of petroleum = Rs 1

Cost of shipping 1L of petroleum = Rs 1/10

Thus, absolute transportation cost is given by,

    \[\begin{array}{*{35}{l}} <!-- /wp:paragraph --> <!-- wp:paragraph --> z\text{ }=\text{ }\left( 7/10 \right)\text{ }x\text{ }+\text{ }\left( 6/10 \right)\text{ }y\text{ }+\text{ }3/10\text{ }\left( 7000\text{ }\text{ }x\text{ }\text{ }y \right)\text{ }+\text{ }3/10\text{ }\left( 4500\text{ }\text{ }x \right)\text{ }+\text{ }4/10\text{ }\left( 3000\text{ }\text{ }y \right)\text{ }+\text{ }2/10\text{ }\left( x\text{ }+\text{ }y\text{ }\text{ }3500 \right) \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> ~ \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \end{array}\]

    \[=\text{ }0.3x\text{ }+\text{ }0.1y\text{ }+\text{ }3950\]

The issue can be figured as given underneath

    \[Limit\text{ }z\text{ }=\text{ }0.3x\text{ }+\text{ }0.1y\text{ }+\text{ }3950\text{ }\ldots \text{ }\ldots \text{ }.\text{ }\left( I \right)\]

Subject to requirements,

    \[x\text{ }+\text{ }y\text{ }\le \text{ }7000\text{ }\ldots \text{ }\ldots \text{ }.\text{ }\left( ii \right)\]

    \[x\text{ }\le \text{ }4500\text{ }\ldots \text{ }\ldots \text{ }..\text{ }\left( iii \right)\]

    \[y\text{ }\le \text{ }3000\text{ }\ldots \text{ }.\text{ }\left( iv \right)\]

    \[\begin{array}{*{35}{l}} <!-- /wp:paragraph --> <!-- wp:paragraph --> x\text{ }+\text{ }y\text{ }\ge \text{ }3500\text{ }\ldots \text{ }\ldots \text{ }\left( v \right) \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> ~ \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \end{array}\]

    \[x,\text{ }y\text{ }\ge \text{ }0\text{ }\ldots \text{ }\ldots \text{ }\left( vi \right)\]

The attainable area controlled by the limitations is given beneath

NCERT Solutions Mathematics Class 12 Chapter 12 - 42

A (3500, 0), B (4500, 0), C (4500, 2500), D (4000, 3000) and E (500, 3000) are the corner points of the attainable area.

The upsides of z at these corner focuses are given beneath

Corner point      

    \[z\text{ }=\text{ }0.3x\text{ }+\text{ }0.1y\text{ }+\text{ }3950\]

 

A (3500, 0)          5000

B (4500, 0)          5300

C (4500, 2500)   5550

D (4000, 3000)   5450

E (500, 3000)      4400      Minimum

The base worth of z is 4400 at (500, 3000)

Thus, the oil provided from warehouse An is 500 L, 3000 L and 3500 L and from terminal B is 4000 L, 0 L and 0 L to petroleum siphons D, E and F separately. Along these lines, the base transportation cost is Rs 4400.

i

More Material